Unformatted text preview: line and a point which is not on that line. Find the line perpendicular to the given
line which passes through the given point.
(a) y = 3 x + 2, P (0, 0) (b) y = −6x + 5, P (3, 2) 13. We shall now prove that y = m1 x + b1 is perpendicular to y = m2 x + b2 if and only if
m1 · m2 = −1. To make our lives easier we shall assume that m1 > 0 and m2 < 0. We can
also “move” the lines so that their point of intersection is the origin without messing things
up, so we’ll assume b1 = b2 = 0. (Take a moment with your classmates to discuss why this is
okay.) Graphing the lines and plotting the points O(0, 0) , P (1, m1 ) and Q(1, m2 ) gives us
the following set up.
y P x O Q The line y = m1 x will be perpendicular to the line y = m2 x if and only if OP Q is a right
triangle. Let d1 be the distance from O to P , let d2 be the distance from O to Q and let d3
be the distance from P to Q. Use the Pythagorean Theorem to show that OP Q is a right
triangle if and only if m1 · m2 = −1 by showing d2 + d2 = d2 if and only if m1 · m2 = −1.
3 2.1 Linear Functions 125 14. Show that if a = b, the line containing the points (a, b) and (...
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